Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 20 Jul 2005 13:57 Daryl McCullough said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > >>> I offered, and you saw, a deductive proof that proves that the > >>> largest natural in a set must be at least as large as the set size. > > Yes. So we have: > > If S is a set of natural numbers, and S has a largest member N, > then N >= the cardinality of S. > > How do you prove that every set of natural numbers has a largest > member? > > -- > Daryl McCullough > Ithaca, NY > > Essentially, the proof shows that no set can have a larger number of naturals in it than the values of all the naturals in it. I will devise a proof without a largest element, if need be, but that "largest element" argument is a waste of time. I still don't see how you can say that for all finite sets this is true, but that one can get an infinite set, and still have all finite numbers. If each finite n in N is the size of the set including all m<=n, then each of them corresponds to a finite set. How do we get an infinite set, then, if m<=n is finite for any finite n in N? -- Smiles, Tony
From: David Kastrup on 20 Jul 2005 13:55 stevendaryl3016(a)yahoo.com (Daryl McCullough) writes: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >>> If TO's assumprtions were actually the case, there would have to be a >>> finite natural so large that adding 1 to it would produce an infinite >>> natural. But TO cannot produce either a largest finite nor a smallest >>> infinite, so the set of all finite naturals is already big enough. >>> >>We have been through all this before. You lay these requirement on >>me, but when I say you cannot have a smallest infinite omega > > But the Peano axioms say that *all* nonempty sets of naturals have a > smallest element. Hey, they say nothing of that sort. Let's see whether we can see that they _imply_ that. But before we can do that, we have to _define_ smallest element. One reasonably natural definition would be: In a subset Q of the naturals N, we call m the smallest element of Q if any set Q' that contains m as an element, and that contains for each of its members k its successor S(k), will contain every member of Q. So we need to show that this property holds for one and only one member of any non-empty subset Q of N. Now if 0 is an element of Q, m=0 obviously leads to Q'=N, Q is a subset of N, and so m=0 obeys the criterion. Now m is not the successor of any number, so if any different m' would also be a minimum, any set containing m' and for each of its members containing its successor would need to contain 0. Since m' is not equal to 0, this rule does not require 0 to be included in Q'', and so can't be minimal. Now let us assume that all numbers "up to" k are not a member of Q. Oh wait, we have to define "up to". You see the kind of brunt work you have to go through before the Peano axiom spew out that a minimum is contained in any subset of the naturals? Of course it is possible, but it is far from self-evident. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Virgil on 20 Jul 2005 13:58 In article <MPG.1d4825defff809ee989f39(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > I proved it for all n in N, which I think you agree is an infinite number of > naturals. But that does not prove that what is true for members of N is true for N itself. Consider the set containing a single apple. By TO's argument, that set must itself be an apple, since that is true for all its members. To uses a similar argument in reverse, that an infinite set must cinatin infinite objects, so not that appple must be a set as well as the set being an apple, at least by TO_logic. > I think you were perhaps one of those claiming that inductive proof > only works for finite iterations, but then it wouldn't work for the infinite > set of naturals, now, would it? I am not going in these stupid circles with > you. TO makes up his own stupid circles, and goes round and round them endlessly. > > Again this goes back to the Cantorian mantra, "no largest finite" It is not strictly Cantorian. To the best of my knowledge, even those mathematicians who avoid infinities do not object to 'no largest finite natural', though all mathematicians object to infinite naturals, at least in standad models. If TO wants infinite naturals, he had better look up Abraham Robinson, et al, but if he thinks Cantorian math is confusing, he will not get far with non-standard analysis.
From: Tony Orlow on 20 Jul 2005 14:02 Daryl McCullough said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > >>> There is nothing in Peano's axioms that states explicitly that all > >>> natural numbers are finite. > > Let's get more specific, and consider the sets S_n = the set of all > natural numbers less than n. Are you claiming that there is a natural > number n such that S_n is not finite? Actually, no, I am saying that for all finite n, S_n is finite, and that, if all n in N are finite, then so is N. Conversely, since S_n is finite for finite n, if S_n is infinite then n is infinite. > > What definition of finite are you using? Less than any infinite number. Not infinite. With a known end, or bound. I am sure you know what I mean. > > -- > Daryl McCullough > Ithaca, NY > > By the way, I don't suppose you would be going to GrassRoots this weekend? -- Smiles, Tony
From: Virgil on 20 Jul 2005 14:01
In article <MPG.1d482eed2a111a4c989f3c(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Whatever you "got", you didn't catch it from me. Fortunately, ost people are immune to the sort of insanity TO has caught. |